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Sagot :
First we find the third side: 26 = X + 8 + 12 --> X = 6. Then, we find (S) to be used in finding the area of the triangle: S = (1/2)(8 + 12 + 6) = 13. To find the area, we solve: A = sqrt (13)(13 - 8)(13 - 12)(13 - 6) = 21.33 cm^2.
[tex]Two \ sides \ of \ a \ triangle \ measures : \\a= 8\ cm, \ \ b= 12\ cm , \ \ c=? \\ perimeter : \ P=26 \ cm \\\\P=a+b+c\\ \\26 =8+12+c \\26=20+c\\c=26-20\\c=6\ cm[/tex]
[tex]Heron's \ formula \\ \\ You \ can \ use \ this \ formula \ to \ find \ the \ area \ of \ a \ triangle \ using \ the \ 3 \ side \ lengths. \\ \\ A = \sqrt{s(s-a)(s-b)(s-c)} \\ \\where \ " s" \ is \ the semi \ perimeter \ of \ the t\ riangle : \\ \\s=\frac{a+b+c}{2}.[/tex]
[tex]s=\frac{8+12+6}{2}=\frac{26}{2}=13 \ cm\\ \\ A = \sqrt{13(13-8)(13-12)(13-6)}\\\\A = \sqrt{13 \cdot 5\cdot 1\cdot 7} \\\\A=\sqrt{455}\ cm^2\\ \\A\approx 21,33 \ cm^2[/tex]
[tex]Heron's \ formula \\ \\ You \ can \ use \ this \ formula \ to \ find \ the \ area \ of \ a \ triangle \ using \ the \ 3 \ side \ lengths. \\ \\ A = \sqrt{s(s-a)(s-b)(s-c)} \\ \\where \ " s" \ is \ the semi \ perimeter \ of \ the t\ riangle : \\ \\s=\frac{a+b+c}{2}.[/tex]
[tex]s=\frac{8+12+6}{2}=\frac{26}{2}=13 \ cm\\ \\ A = \sqrt{13(13-8)(13-12)(13-6)}\\\\A = \sqrt{13 \cdot 5\cdot 1\cdot 7} \\\\A=\sqrt{455}\ cm^2\\ \\A\approx 21,33 \ cm^2[/tex]
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